What is area y= 1/2 x,y= 6/x ,y=2x ?

The answer to area y= 1/2 x,y= 6/x ,y=2x is 12ln(2)

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area y= 1/2 x,y= 6/x ,y=2x

area

y = \frac{1}{2} x, y = \frac{6}{x}, y = 2 x

12 ln (2)

Decimal Notation

8.31776 \dots

Solution steps

To find the intersection points solve

f_{i} (x) = f_{j} (x)

\frac{1}{2} x = \frac{6}{x} : x = 23, x = - 23

\frac{1}{2} x = 2 x : x = 0

\frac{6}{x} = 2 x : x = 3, x = - 3

= \int_{- 23}^{- 3} ∣ f_{1} (x) - f_{2} (x) ∣ d x + \int_{- 3}^{0} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{0}^{3} ∣ f_{1} (x) - f_{3} (x) ∣ d x + \int_{3}^{23} ∣ f_{2} (x) - f_{1} (x) ∣ d x

= \int_{- 23}^{- 3} \frac{1}{2} x - \frac{6}{x} d x + \int_{- 3}^{0} \frac{1}{2} x - 2 x d x + \int_{0}^{3} \frac{1}{2} x - 2 x d x + \int_{3}^{23} \frac{6}{x} - \frac{1}{2} x d x

\int_{- 23}^{- 3} \frac{1}{2} x - \frac{6}{x} d x = - \frac{9}{4} + 6 ln (2)

\int_{- 3}^{0} \frac{1}{2} x - 2 x d x = \frac{9}{4}

\int_{0}^{3} \frac{1}{2} x - 2 x d x = \frac{9}{4}

\int_{3}^{23} \frac{6}{x} - \frac{1}{2} x d x = 6 ln (2) - \frac{9}{4}

= - \frac{9}{4} + 6 ln (2) + \frac{9}{4} + \frac{9}{4} + 6 ln (2) - \frac{9}{4}

Simplify

= 12 ln (2)

x \to 5 - lim (\frac{3}{x - 5})

x \to 2 - lim (x - 1)

- 4 \cdot sec^{2} (4 x)

f (x) = \frac{cos ( x )}{1 - sin ( x )}

\int \frac{2 x ^{3} + 3 x ^{2} - 4}{x ^{2} - 4 x + 3} d x

What is area y= 1/2 x,y= 6/x ,y=2x ?
The answer to area y= 1/2 x,y= 6/x ,y=2x is 12ln(2)